\(\int \frac {(-5+10 x-x^2-2 x^3) \log (x)+(5-x^2) \log (\frac {5-x^2}{x})}{-10 x+2 x^3} \, dx\) [9936]

Optimal result

Integrand size = 51, antiderivative size = 29 \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=-4+x-\log (x)+\left (1-x-\frac {1}{2} \log \left (\frac {5}{x}-x\right )\right ) \log (x) \]

[Out]

x-ln(x)-4+(1-x-1/2*ln(5/x-x))*ln(x)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 15, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.157, Rules used = {1607, 6857, 2404, 2332, 2338, 2375, 2438, 2604} \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )+x+x (-\log (x)) \]

[In]

Int[((-5 + 10*x - x^2 - 2*x^3)*Log[x] + (5 - x^2)*Log[(5 - x^2)/x])/(-10*x + 2*x^3),x]

[Out]

x - x*Log[x] - (Log[x]*Log[(5 - x^2)/x])/2

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2375

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[f^m*Log[1 + e*(x^r/d)]*((a + b*Log[c*x^n])^p/(e*r)), x] - Dist[b*f^m*n*(p/(e*r)), Int[Log[1 + e*(x^r/d)]*((
a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

Rule 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2604

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[Log[d + e*x]*((a + b
*Log[c*RFx^p])^n/e), x] - Dist[b*n*(p/e), Int[Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{x \left (-10+2 x^2\right )} \, dx \\ & = \int \left (-\frac {\left (5-10 x+x^2+2 x^3\right ) \log (x)}{2 x \left (-5+x^2\right )}-\frac {\log \left (\frac {5-x^2}{x}\right )}{2 x}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {\left (5-10 x+x^2+2 x^3\right ) \log (x)}{x \left (-5+x^2\right )} \, dx\right )-\frac {1}{2} \int \frac {\log \left (\frac {5-x^2}{x}\right )}{x} \, dx \\ & = -\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )+\frac {1}{2} \int \frac {x \left (-2-\frac {5-x^2}{x^2}\right ) \log (x)}{5-x^2} \, dx-\frac {1}{2} \int \left (2 \log (x)-\frac {\log (x)}{x}+\frac {2 x \log (x)}{-5+x^2}\right ) \, dx \\ & = -\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )+\frac {1}{2} \int \frac {\log (x)}{x} \, dx+\frac {1}{2} \int \left (-\frac {\log (x)}{x}+\frac {2 x \log (x)}{-5+x^2}\right ) \, dx-\int \log (x) \, dx-\int \frac {x \log (x)}{-5+x^2} \, dx \\ & = x-x \log (x)+\frac {\log ^2(x)}{4}-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )-\frac {1}{2} \log (x) \log \left (1-\frac {x^2}{5}\right )-\frac {1}{2} \int \frac {\log (x)}{x} \, dx+\frac {1}{2} \int \frac {\log \left (1-\frac {x^2}{5}\right )}{x} \, dx+\int \frac {x \log (x)}{-5+x^2} \, dx \\ & = x-x \log (x)-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right )-\frac {\operatorname {PolyLog}\left (2,\frac {x^2}{5}\right )}{4}-\frac {1}{2} \int \frac {\log \left (1-\frac {x^2}{5}\right )}{x} \, dx \\ & = x-x \log (x)-\frac {1}{2} \log (x) \log \left (\frac {5-x^2}{x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=\frac {1}{2} \left (2 x-2 x \log (x)-\log (x) \log \left (\frac {5-x^2}{x}\right )\right ) \]

[In]

Integrate[((-5 + 10*x - x^2 - 2*x^3)*Log[x] + (5 - x^2)*Log[(5 - x^2)/x])/(-10*x + 2*x^3),x]

[Out]

(2*x - 2*x*Log[x] - Log[x]*Log[(5 - x^2)/x])/2

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79

method	result	size
parallelrisch	\(-x \ln \left (x \right )-\frac {\ln \left (x \right ) \ln \left (-\frac {x^{2}-5}{x}\right )}{2}+x\)	\(23\)
default	\(-\frac {\ln \left (x \right ) \ln \left (\frac {-x^{2}+5}{x}\right )}{2}-x \ln \left (x \right )+x\)	\(24\)
parts	\(-\frac {\ln \left (x \right ) \ln \left (\frac {-x^{2}+5}{x}\right )}{2}-x \ln \left (x \right )+x\)	\(24\)
risch	\(-\frac {\ln \left (x^{2}-5\right ) \ln \left (x \right )}{2}+\frac {\ln \left (x \right )^{2}}{2}-x \ln \left (x \right )+x +\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )}^{2}}{2}-\frac {i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \left (x^{2}-5\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )}^{2}}{4}+\frac {i \ln \left (x \right ) \pi \,\operatorname {csgn}\left (i \left (x^{2}-5\right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )}{4}-\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )}^{3}}{4}-\frac {i \ln \left (x \right ) \pi {\operatorname {csgn}\left (\frac {i \left (x^{2}-5\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{4}-\frac {i \ln \left (x \right ) \pi }{2}\)	\(160\)

[In]

int(((-2*x^3-x^2+10*x-5)*ln(x)+(-x^2+5)*ln((-x^2+5)/x))/(2*x^3-10*x),x,method=_RETURNVERBOSE)

[Out]

-x*ln(x)-1/2*ln(x)*ln(-(x^2-5)/x)+x

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=-\frac {1}{2} \, {\left (2 \, x + \log \left (-\frac {x^{2} - 5}{x}\right )\right )} \log \left (x\right ) + x \]

[In]

integrate(((-2*x^3-x^2+10*x-5)*log(x)+(-x^2+5)*log((-x^2+5)/x))/(2*x^3-10*x),x, algorithm="fricas")

[Out]

-1/2*(2*x + log(-(x^2 - 5)/x))*log(x) + x

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=- x \log {\left (x \right )} + x - \frac {\log {\left (x \right )} \log {\left (\frac {5 - x^{2}}{x} \right )}}{2} \]

[In]

integrate(((-2*x**3-x**2+10*x-5)*ln(x)+(-x**2+5)*ln((-x**2+5)/x))/(2*x**3-10*x),x)

[Out]

-x*log(x) + x - log(x)*log((5 - x**2)/x)/2

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=-x \log \left (x\right ) - \frac {1}{2} \, \log \left (-x^{2} + 5\right ) \log \left (x\right ) + \frac {1}{2} \, \log \left (x\right )^{2} + x \]

[In]

integrate(((-2*x^3-x^2+10*x-5)*log(x)+(-x^2+5)*log((-x^2+5)/x))/(2*x^3-10*x),x, algorithm="maxima")

[Out]

-x*log(x) - 1/2*log(-x^2 + 5)*log(x) + 1/2*log(x)^2 + x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=-x \log \left (x\right ) - \frac {1}{2} \, \log \left (-x^{2} + 5\right ) \log \left (x\right ) + \frac {1}{2} \, \log \left (x\right )^{2} + x \]

[In]

integrate(((-2*x^3-x^2+10*x-5)*log(x)+(-x^2+5)*log((-x^2+5)/x))/(2*x^3-10*x),x, algorithm="giac")

[Out]

-x*log(x) - 1/2*log(-x^2 + 5)*log(x) + 1/2*log(x)^2 + x

Mupad [B] (verification not implemented)

Time = 16.42 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\left (-5+10 x-x^2-2 x^3\right ) \log (x)+\left (5-x^2\right ) \log \left (\frac {5-x^2}{x}\right )}{-10 x+2 x^3} \, dx=x-\frac {\ln \left (-\frac {x^2-5}{x}\right )\,\ln \left (x\right )}{2}-x\,\ln \left (x\right ) \]

[In]

int((log(-(x^2 - 5)/x)*(x^2 - 5) + log(x)*(x^2 - 10*x + 2*x^3 + 5))/(10*x - 2*x^3),x)

[Out]

x - (log(-(x^2 - 5)/x)*log(x))/2 - x*log(x)